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Hunter-Best [27]
3 years ago
9

The prime factorization of a number is 3^2x5^3x7. Which statement is true about the factors of the number? ? Twenty-one is a fac

tor of the number because both 3 and 7 are prime factors. Twenty-one is not a factor of the number because 21 is not prime. Ninety is a factor of the number because 3^2=9 and 90 is divisible by 9. Ninety is not a factor of the number because 90 is not divisible by 7.
Mathematics
2 answers:
rodikova [14]3 years ago
7 0
The 1st statement is true.

21 is a factor because both 3 and 7 are factors.
olga_2 [115]3 years ago
6 0

Answer:

Twenty-one is a factor of the number because both 3 and 7 are prime factors.

Step-by-step explanation:

Given number is,

3^2\times 5^3\times 7

=3\times 3\times 5\times 5\times 5\times 7

Where, 3, 5 and 7 are prime numbers ( only divisible by 1 and itself ),

⇒ Both 3 and 7 are prime factors of the given number,

⇒ 21 is a factor of the given number.

Thus, first option is correct.

⇒ Second option is incorrect.

Now, 5 is factor of the given number but 2 is not,

⇒ 10 is not a factor of the given number,

⇒ 90 is not a factor of the given number,

⇒ Third option is incorrect.

Suppose 90 is divisible by 7,

⇒ 90 = 7a

Where a is any whole number,

⇒ 7=\frac{90}{a}

3^2\times 5^3\times 7=3^2\times 5^3\times \frac{90}{a}

Since, 90 could be a factor of this number, if a = 3 or 5 or their multiple,

For the other values of a, 90 can not be the factor,

Hence, there is no effect of divisibility of 90 by 7 on having 90 as a factor of the given number,

⇒ Fourth option is incorrect.

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Answer:

B

Step-by-step explanation:

the absolute value of -13 is 13, and 13 is greater than 5

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3 years ago
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Talja [164]

Answer:

b

Step-by-step explanation:

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3 years ago
rectangle ABCD is graphed in the cordinate plane. The following are the verticies of the rectangle; A(-6,-4),B(-4,-4),C(-4,-2),
Sholpan [36]

Answer:

Therefore Perimeter of Rectangle ABCD is 4 units

Step-by-step explanation:

Given:

ABCD is a Rectangle.

A(-6,-4),

B(-4,-4),

C(-4,-2), and

D (-6,-2).

To Find :

Perimeter of Rectangle = ?

Solution:

Perimeter of Rectangle is given as

\textrm{Perimeter of Rectangle}=2(Length+Width)

Length = AB

Width = BC

Now By Distance Formula  we have'

l(AB) = \sqrt{((x_{2}-x_{1})^{2}+(y_{2}-y_{1})^{2} )}

Substituting the values we get

l(AB) = \sqrt{((-4-(-6))^{2}+(-4-(-4))^{2} )}

l(AB) = \sqrt{((2)^{2}+(0)^{2} )}=2\ unit

Similarly

l(BC) = \sqrt{((-4-(-4))^{2}+(-2-(-4))^{2} )}

l(BC) = \sqrt{((0)^{2}+(2)^{2} )}=2\ unit

Therefore now

Length = AB = 2 unit

Width = BC = 2 unit

Substituting the values  in Perimeter we get

\textrm{Perimeter of Rectangle}=2(2+2)=2(4)=8\ unit

Therefore Perimeter of Rectangle ABCD is 4 units

8 0
3 years ago
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(n-10)

Step-by-step explanation:

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2 years ago
What are the solution(s) to the quadratic equation 40 − x2 = 0? x = ±2 x = ±4 x = ±2 x = ±4
hichkok12 [17]

Answer: x=\±2\sqrt{10}

Step-by-step explanation:

To find the solutions of the quadratic equation given in the problem, you can apply the proccedure shown below:

- Subtract 40 from both sides of the equation.

- Multiply both sides of the equation by -1

- Apply square root to both sides of the equation.

Therefore, by applyin the steps above, you obtain:

40-x^2-40=0-40\\-x^2=-40\\(-1)(-x^2)=(-40)(-1)\\\\x^2=40\\\\\sqrt{x^2}=\±\sqrt{40}\\\\x=\±2\sqrt{10}

3 0
3 years ago
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